Question

A random sample of 28 policies form insurance company A shows an average net profit of $39 per policy, with a standard deviation of $3.70 per policy. A random sample of 34 policies from insurance company B shows an average net profit of $47 per policy with a standard deviation of $3.90 per policy. Assume the policy profits are normally distributed and have unequal variances.   Use a 0.01 level of significance to test whether the profits per policy are significantly greater for company B than for company A.

Answer 1

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 <   0          
                  
Level of Significance ,    α =    0.01          
                  
Sample #1   ---->   sample 1          
mean of sample 1,    x̅1=   39.00          
standard deviation of sample 1,   s1 =    3.7          
size of sample 1,    n1=   28          
                  
Sample #2   ---->   sample 2          
mean of sample 2,    x̅2=   47.000          
standard deviation of sample 2,   s2 =    3.90          
size of sample 2,    n2=   34          
                  
difference in sample means = x̅1-x̅2 =    39.000   -   47.0000   =   -8.0000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.9676          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -8.0000   /   0.9676   ) =   -8.2677
                  

                  
  
p-value =        0.0000   [ excel function: =T.DIST(t stat,df) ]       
Conclusion:     p-value<α , Reject null hypothesis  
           
There is enough evidence to say that the profits per policy are significantly greater for company B than for company A.

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